Monday, September 24, 2018

Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 32

Find the integrals 913x2xdx

3x2xdx=(3xx2x)dx3x2xdx=(3x122x12)dx3x2xdx=3x12dx2x12dx3x2xdx=3(x12+112+1)2(x12+112+1)+C3x2xdx=2x322(2x12)+C3x2xdx=2x324x12+C913x2xdx=2(9)324(9)12+C[2(1)324(1)12+C]913x2xdx=2[(9)12]34(3)+C2+4C913x2xdx=2(3)312+2913x2xdx=5412+2913x2xdx=44

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